Let a₁, a₂, be an arithmetic progression of positive integers with common difference d . Suppose the terms a₂, a₅, and a₁₄ form a geometric progression. If the sum of the infinite geometric progression with first term a₁ and common ratio 1 d lies strictly in the interval (4, 5) , then the 10^ th term of the arithmetic progression is equal to

Step-by-step solution

Let the first term of the arithmetic progression be a and the common difference be d , where a, d Z ^+ . The terms a₂, a₅, and a₁₄ are a+d, a+4d, and a+13d respectively. Since they form a geometric progression: (a+4d)^2 = (a+d)(a+13d) a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2 3d^2 = 6ad Since d > 0 , we can divide by 3d to get d = 2a . The infinite geometric progression has first term a₁ = a and common ratio r = 1 d = 1 2a . The sum of this infinite GP is: S_ = a 1 - 1 2a = 2a^2 2a - 1 We are given that 4 4 Since a is